Question

The variation of electric field between two charges $q_1$ and $q_2$ along the line joining the charges is plotted against distance from $q_1$ (taking rightward direction of electric field as positive) as shown in the figure. Then, the correct statement is

• A. $q_1$ and $q_2$ are positive and $q_1<q_2$
• B. $q_1$ and $q_2$ are positive and $q_1>q_2$
• C. $q_1$ is positive and $q_2$ is negative ; $q_1<\left|q_2\right|$
• D. $q_1$ and $q_2$ are negative and $\left|q_1\right|<\left|q_2\right|$

Answer 1

The correct option is A $q_1$ and $q_2$ are positive and $q_1<q_2$


We can see from the graph that:

$\left(1\right).$
Just to the right of $q_1$, the electric field is $+\alpha$ (let) or in positive direction (away from $q_1$).
Hence, $q_1$ is positive because $r$ and $E$ both are positive.

$\left(2\right).$
Just to the left of $q_2$, the electric field is $-\alpha$ (let) or toward left (away from $q_2$).
Hence, $q_2$ is also positive because $r$ and $E$ both are negative.

$\left(3\right).$
$E=0$ is near $q_1$.
Hence $q_1<q_2$ because $E\propto q$ and $E\propto \frac{1}{r}$.

Hence, option $\left(a\right)$ is the correct answer.


Why this question?
Concept - For positive charge in $E-x$ curve, $x-$ positive, $E$ positive and $x$ negative, $E$ negative.
For negative charge in $E-x$ curve,
$x-$ positive, $E$ negative and $x$ negative, $E$ positive.